Solution
= Solution
Set $t=x^2$, so $dx=dt/(2\sqrt t)$. Then
$$
I=\frac12\int_0^1t^{-1/3}(1-t)^{1/3}dt
=\frac12B(2/3,4/3).
$$
The <beta--gamma identity>, the <Gamma function recurrence> and the <gamma reflection formula> give
$$
I=\frac12\Gamma(2/3)\Gamma(4/3)
=\frac16\Gamma(2/3)\Gamma(1/3)
=\boxed{\frac{\pi}{6\sin(\pi/3)}}.
$$